133 (2008)
MATHEMATICA BOHEMICA
No. 4, 377–387
TRIBONACCI MODULO 2t AND 11t
Jiří Klaška, Brno
(Received May 31, 2007)
Abstract. Our previous research was devoted to the problem of determining the primitive
∞
periods of the sequences (Gn mod pt )∞
n=1 where (Gn )n=1 is a Tribonacci sequence defined
by an arbitrary triple of integers. The solution to this problem was found for the case of
powers of an arbitrary prime p 6= 2, 11. In this paper, which could be seen as a completion
of our preceding investigation, we find solution for the case of singular primes p = 2, 11.
Keywords: Tribonacci, modular periodicity, periodic sequence
MSC 2000 : 11B50, 11B39
1. Introduction
Having a linear recurrence formula of order k with integer coefficients we can
construct the corresponding characteristic polynomial f (x). If f (x) has no multiple
roots then its discriminant is a non zero integer and so it is divisible by only a finite
number of prime divisors. When investigating modular periodicity of the sequences
defined by these formulas, the primes that divide the discriminant of f (x) form exceptions and have to be considered separately. The exceptional primes p correspond
to the cases of f (x) having multiple roots over the field F p = Z/pZ of residue classes
modulo p. In this paper, which could be seen as an extension of our previous paper
[1], we focus on the Tribonacci case. It is well known, see for example [2, p. 310],
that the primes p = 2, 11 are the only primes for which the Tribonacci characteristic
polynomial g(x) = x3 − x2 − x − 1 has multiple roots.
Let us now review the notation introduced in [1]. Let (gn )∞
n=1 denote a Tribonacci
sequence defined by the recurrence formula gn+3 = gn+2 + gn+1 + gn and the triple of
initial values [0, 0, 1]. Let further (Gn )∞
n=1 denote the generalized Tribonacci sequence
defined by an arbitrary triple [a, b, c] of integers. We will denote the primitive peri∞
ods of the sequences (gn mod m)∞
n=1 and (Gn mod m)n=1 by h(m) and h(m)[a, b, c]
377
respectively. In 1978, M. E. Waddill [3, Theorem 8] proved that for any prime p and
t ∈ N = {1, 2, 3, . . .}, we have:
If h(p) 6= h(p2 ), then h(pt ) = pt−1 h(p).
(1.1)
This paper aims at determining the numbers h(pt )[a, b, c] and find the relationships
between h(pt )[a, b, c] and h(p)[a, b, c] for the primes p = 2, 11. The case of p 6= 2, 11
is solved in [1]. The methods used in proofs of this paper will mostly be based on
matrix algebra. As usual, by T we will denote the Tribonacci matrix
(1.2)
0 1

T = 0 0
1 1


0
1
1

gn
and T n =  gn+1
gn+2
gn−1 + gn
gn + gn+1
gn+1 + gn+2

gn+1
gn+2  for n > 1.
gn+3
Put x0 = [a, b, c]τ and xn = [Gn+1 , Gn+2 , Gn+3 ]τ where τ denotes transposition.
Then the triple xn may be expressed by means of x0 as follows: xn = T n x0 . Thus
the primitive period of the sequence (Gn mod m)∞
n=1 defined by a triple [a, b, c] for
an arbitrary module m > 1 is equal to the smallest number h for which T h x0 ≡ x0
(mod m). By [1, Lemma 2.1], the investigation of the primitive periods of Tribonacci
sequences modulo pt is restricted to sequences beginning with the triples [a, b, c] 6≡
[0, 0, 0] (mod p). In the opposite case, for any t ∈ N and 1 6 i 6 t, we have
h(pt )[pt−i a, pt−i b, pt−i c] = h(pi )[a, b, c]. For this reason, we will investigate only the
triples satisfying [a, b, c] 6≡ [0, 0, 0] (mod p).
2. Tribonacci modulo 2t
We can easily calculate h(2) = 4 and h(22 ) = 8. By (1.1) we have h(2t ) =
2 h(2) = 2t+1 and so h(2t )[a, b, c] | 2t+1 for any [a, b, c]. For p = 2, the multiplicity
of the root α = 1 of the polynomial g(x) is greater than char(F 2 ) = 2 and therefore
2
(Gn mod 2)∞
n=1 cannot be expressed as Gn mod 2 = c1 + c2 n + c3 n as usual. The
∞
2 ∞
sequences (1)∞
n=1 , (n)n=1 , (n )n=1 are dependent over F 2 and do not form a basis.
Despite that, for some triples [a, b, c] 6≡ [0, 0, 0] (mod 2), the numbers h(2t )[a, b, c] can
be determined using the results derived in [1]. In the first place, it is proved in [1,
Theorem 3.1] that, if (D(a, b, c), m) = 1 where D(a, b, c) is a cubic form defined by
t−1
(2.1)
D(a, b, c) = a3 + 2b3 + c3 − 2abc + 2a2 b + 2ab2 − 2bc2 + a2 c − ac2 ,
then h(m)[a, b, c] = h(m) for any modulus m > 1. The following theorem is an easy
consequence of the above assertions.
378
Theorem 2.1. If D(a, b, c) is an odd number, then h(2t )[a, b, c] = h(2t ) = 2t+1 .
Hence, we have h(2t )[a, b, c] = 2t−1 · h(2)[a, b, c].
It is easy to verify that the premise of Theorem 2.1 is true if and only if [a, b, c]
is congruent modulo 2 with some of the triples [0, 0, 1], [1, 0, 0], [1, 1, 0], [0, 1, 1].
Therefore it suffices to investigate the cases of the triple [a, b, c] being congruent
modulo 2 with some of the triples [0, 1, 0], [1, 0, 1], [1, 1, 1]. The following assertions
will be important for the proofs of the main theorems 2.4, 2.5 and 2.6.
Lemma 2.2. For any modulus of the form 2t where t > 5, the following congruences hold:
g2t−1 −1 ≡ −1 (mod 2t ),
g2t−1 ≡ 2t−2 + 1 (mod 2t ),
g2t−1 +1 ≡ 0 (mod 2t ),
g2t−1 +2 ≡ 2t−2 (mod 2t ),
g2t−1 +3 ≡ 2t−1 + 1 (mod 2t ).
P r o o f . Using methods of matrix algebra, we will prove all the congruences in
(2.2) simultaneously. Let us consider a Tribonacci matrix T . Due to (1.2), it suffices
to prove that for any t > 5 we have
t−1
T2

2t−2 +1 2t−2
0

≡0
2t−2 +1 2t−2
t−2
t−2
t−1
2
2
2 +1

1 1
t−2
t

≡ E + 2 A (mod 2 ), where A = 0 1
1 1


0
1
2
and E is an identity matrix. Let us first prove the congruence for t = 5. By direct
calculation, we can verify that
4
T2
1705
=  3136
5768

2632
4841
8904
  3
3136
2 +1
23
5768  ≡  0
23 + 1
3
10609
2
23

0
23  (mod 25 ).
24 + 1
Let us further assume that the congruence holds for t > 5. Since AE = EA, we have
t
T 2 ≡ (E + 2t−2 A)2 ≡ E + 2t−1 A (mod 2t+1 ), which proves (2.2).
379
Consequence 2.3.
congruences hold:
For any modulus of the form 2t where t > 3, the following
g2t −1 ≡ −1 (mod 2t ), g2t ≡ 2t−1 + 1 (mod 2t ),
g2t +1 ≡ 0 (mod 2t ), g2t +2 ≡ 2t−1 (mod 2t ),
(2.3)
g2t +3 ≡ 1 (mod 2t ).
P r o o f . For t = 3, (2.3) can be verified by direct calculation. For t > 4, (2.3)
follows from (2.2).
Theorem 2.4. If [a, b, c] ≡ [0, 1, 0] (mod 2), then for t > 1 we have
h(2t )[a, b, c] = 2t+1 .
(2.4)
P r o o f . Clearly, it is sufficient to prove that x2t 6≡ x0 (mod 2t ), that is, that 2t
is not a period. The triple [a, b, c] can be written as x0 = [2a1 , 1 + 2b1 , 2c1 ]τ where
a1 , b1 , c1 ∈ Z. For t = 2 we have

 


2a1
2 + 2a1
1 2 2
2
T 2 x0 =  2 3 4   1 + 2b1  ≡  3 + 2b1  (mod 22 ).
2c1
4 6 7
2 + 2c1
2
Suppose that T 2 x0 ≡ x0 (mod 22 ). Then we have
[2 + 2a1 , 3 + 2b1 , 2 + 2c1 ] ≡ [2a1 , 1 + 2b1 , 2c1 ](mod 22 ).
Hence [2, 3, 2] ≡ [0, 1, 0](mod 22 ),
have
 t−1
2
+ 1 2t−1
t
2

T x0 ≡ 0
2t−1 + 1
2t−1
2t−1
which is a contradiction. If t > 3, then by (2.3) we


 
2a1
2a1 + 2t−1
0
2t−1   1 + 2b1  ≡  1 + 2b1 + 2t−1  (mod 2t ).
2c1 + 2t−1
2c1
1
t
Suppose that T 2 x0 ≡ x0 (mod 2t ). Then we have
[2a1 + 2t−1 , 2t−1 , 2c1 + 2t−1 ] ≡ [2a1 , 1 + 2b1 , 2c1 ] (mod 2t ).
By matching terms, we obtain 2t−1 ≡ 0 (mod 2t ) and thus a contradiction.
It is not difficult to rephrase Theorem 2.4 to include the triples [a, b, c] ≡ [1, 0, 1].
Clearly, there is exactly one triple of the form x0 = [2(c1 − a1 − b1 ), 1 + 2a1 , 2b1 ]τ
corresponding to each triple x1 = [1 + 2a1 , 2b1 , 1 + 2c1 ]τ . Since T x0 = x1 , the triples
x0 and x1 define sequences with identical primitive periods. By 2.4, this primitive
period equals 2t+1 . This proves the following theorem.
380
Theorem 2.5. If [a, b, c] ≡ [1, 0, 1] (mod 2), then for t > 1 we have
h(2t )[a, b, c] = 2t+1 .
(2.5)
We can also use the procedure from 2.4 to prove the following theorem:
Theorem 2.6. If [a, b, c] ≡ [1, 1, 1] (mod 2), then for t > 1 we have
h(2t )[a, b, c] = 2t .
(2.6)
P r o o f . The triple [a, b, c] can be written as x0 = [1+2a1 , 1+2b1 , 1+2c1 ]τ where
t
a1 , b1 , c1 ∈ Z. Suppose t > 5. Then by Lemma 2.2 we have T 2 x0 ≡ x0 (mod 2t )
and so h(2t )[a, b, c] | 2t . It is now sufficient to prove that x2t−1 6≡ x0 (mod 2t ), that
is, that 2t−1 is not a period. By (2.2) we have
x2t−1
2t−2 + 1
2t−1
≡T
x0 ≡  0
2t−2

2t−2
2t−2 + 1
2t−2


0
1 + 2a1
  1 + 2b1  (mod 2t ).
2t−2
2t−1 + 1
1 + 2c1
It follows that
x2t−1 ≡ [1 +2a1 +2t−1 (1+ a1 +b1 ), 1 +2b1 + 2t−1 (1+ b1 + c1 ), 1 +2c1 + 2t−1 (a1 +b1 )]τ .
Suppose x2t−1 ≡ x0 (mod 2t ). Matching the terms yields that
2t−1 (1 + a1 + b1 ) ≡ 0, 2t−1 (1 + b1 + c1 ) ≡ 0, 2t−1 (a1 + b1 ) ≡ 0 (mod 2t ).
Hence 1 ≡ 0 (mod 2) and a contradiction follows. To prove the cases of t = 2, 3, 4 is
easy and can be left to the reader.
R e m a r k 2.7. Theorems 2.4, 2.5, and 2.6 are true for t > 1. In particular, for
t = 1 we have h(2)[1, 1, 1] = 1 and h(2)[0, 1, 0] = h(2)[1, 0, 1] = 2.
Corollary 2.8. If a triple [a, b, c] is congruent modulo 2 with some of the triples
[0, 1, 0], [1, 0, 1], [1, 1, 1], then for any t > 1 we have h(2t )[a, b, c] = 2t · h(2)[a, b, c].
381
3. Tribonacci modulo 11t
The determination of primitive periods modulo 11t will be somewhat more complicated. We can directly verify that h(11) = 110 and h(112 ) = 1210. Now it follows
from (1.1) that h(11t ) = 10 · 11t for any t ∈ N and thus, for any triple [a, b, c], we
have h(11t )[a, b, c] | 10 · 11t . As x3 − x2 − x − 1 ≡ (x − 9)(x − 7)2 (mod 11) and
n ∞
n ∞
(9n )∞
n=1 , (7 )n=1 , (n7 )n=1 are linearly independent over F 11 , we have
Gn ≡ c1 · 9n + (c2 + c3 n) · 7n (mod 11),
(3.1)
where the coefficients c1 , c2 , c3 are uniquely determined by the triple [a, b, c]. Let
ord11 (ε) denote the order of ε 6≡ 0 (mod 11) in the multiplicative group of F 11 . It
is easy to see that ord11 (9) = 5 and ord11 (7) = 10. Now yields (3.1) that for any
[a, b, c] 6≡ [0, 0, 0](mod 11), h(11)[a, b, c] is equal exactly to one of the numbers 5, 10
and 110. This, together with h(11)[a, b, c] | h(11t )[a, b, c], implies that for [a, b, c] 6≡
[0, 0, 0](mod 11) the only forms of the periods h(11t )[a, b, c] are 5·11i and 10·11i where
i ∈ {0, 1, . . . , t}. Consequently, there exists no triple [a, b, c] for which h(11t )[a, b, c] =
2 · 11i . In some cases, h(11t )[a, b, c] can be determined using the form D(a, b, c).
However, there are triples for which h(11t )[a, b, c] = h(11t ) and also D(a, b, c) ≡
0 (mod 11). Thus D(a, b, c) cannot be used to determine all the triples for which
h(11t )[a, b, c] = h(11t ).
Lemma 3.1. Let t > 3 and h = 10 · 11t−2 . Then we have the following congruences:
gh−1 ≡ 25 · 11t−2 − 1 (mod 11t ),
gh ≡ 65 · 11t−2 + 1 (mod 11t ),
gh+1 ≡ 26 · 11t−2 (mod 11t ),
(3.2)
gh+2 ≡ 116 · 11t−2 (mod 11t ),
gh+3 ≡ 86 · 11t−2 + 1 (mod 11t ).
P r o o f . By (1.2), it is sufficient to prove that
t−2
T 10·11
65 · 11t−2 + 1
≡  26 · 11t−2
116 · 11t−2

i.e.
t−2
T 10·11
382
90 · 11t−2
91 · 11t−2 + 1
21 · 11t−2

26 · 11t−2
 (mod 11t ),
116 · 11t−2
t−2
86 · 11
+1
65
t−2
t

≡ E + 11 A (mod 11 ), where A = 26
116

90
91
21

26
116  .
86
In the first induction step, we verify that the congruence is true for t = 3.
T 10·11
716
≡  286
1276

990
1002
231

286
1276  ≡ E + 11A (mod 113 ).
947
Suppose now that the assertion is true for a fixed t > 3 and let us prove it for t + 1.
Since A, E commute, using the binomial expansion we obtain that
T
10·11t−1
t−2
≡ (E + 11
11
A)
11 X
11
(11t−2 A)i
≡
i
i=0
≡ E + 11t−1 A + 5 · 112t−3 A2 (mod 11t+1 )
and A2 ≡ 0 (mod 11) proves (3.2).
Consequence 3.2. Let t > 1 and h = 10 · 11t−1 . Then for any modulus of the
form 11t the following congruences hold:
gh−1 ≡ 3 · 11t−1 − 1 (mod 11t ),
(3.3)
gh ≡ 10 · 11t−1 + 1 (mod 11t ),
gh+1 ≡ 4 · 11t−1 (mod 11t ),
gh+2 ≡ 6 · 11t−1 (mod 11t ),
gh+3 ≡ 9 · 11t−1 + 1 (mod 11t ).
P r o o f . For t = 1, (3.3) can be easily verified by direct calculation. For t > 2,
(3.3) follows from (3.2).
Theorem 3.3. For any t ∈ N we have h(11t )[a, b, c] | 10 · 11t−1 if and only if
c ≡ 3a + 5b (mod 11). Moreover, for any t > 1, if h(11t )[a, b, c] | 10 · 11t−2 then
[a, b, c] ≡ [0, 0, 0] (mod 11).
P r o o f . Let h(11t )[a, b, c] | 10 · 11t−1 . Then (3.3) implies
10 · 11t−1 + 1
 4 · 11t−1
6 · 11t−1

2 · 11t−1
3 · 11t−1 + 1
10 · 11t−1
   
a
a
4 · 11t−1
  b  ≡  b  (mod 11t ).
6 · 11t−1
c
c
9 · 11t−1 + 1
A simple modification of the system yields
   
0
a
10 2 4
 4 3 6   b  ≡  0  (mod 11).
0
c
6 10 9

383
The congruences of this system are linearly dependent over F 11 with the entire system
being equivalent to the single congruence 10a + 2b + 4c ≡ 0 (mod 11). Hence, we
have c ≡ 3a + 5b (mod 11).
Let h(11t )[a, b, c] | 10 · 11t−2 . The validity of the implication for t = 2 is not
difficult to verify by direct calculation. If t > 3, then by (3.2) we have
65 · 11t−2 + 1
 26 · 11t−2
116 · 11t−2
90 · 11t−2
91 · 11t−2 + 1
21 · 11t−2

This system is equivalent to

65 90
 26 91
116 21
   
a
26 · 11t−2
a
t−2




b ≡ b  (mod 11t ).
116 · 11
t−2
86 · 11
+1
c
c
   
0
a
26
116   b  ≡  0  (mod 112 ).
0
c
86
The last system has exactly 121 non-congruent solutions over
written as [11r, 11s, 11(3r + 5s)] where r, s are integers.
/112 Z that can be
Z
R e m a r k 3.4. It follows from 3.3 that, if t > 1 and [a, b, c] 6≡ [0, 0, 0] (mod 11),
then h(11t )[a, b, c] is equal to some of the numbers 5 · 11t−1 , 10 · 11t−1 , 5 · 11t , 10 · 11t .
The following lemmas will help us to determine which of the cases will occur for a
given [a, b, c]. We will also prove that there exists no triple for which h(11t )[a, b, c] =
5 · 11t .
Lemma 3.5. For any t ∈ N we have
t
T 5·11
(3.4)
7 4
≡ A (mod 11) where A =  6 2
10 5


6
10  .
1
Moreover, A2t ≡ E (mod 11).
P r o o f . For t = 1, (3.4) is true since
35731770264967 55158741162067
55 
T = 65720971788709 101452742053676
120879712950776 186600684739485

 
7
65720971788709


120879712950776 ≡ 6
10
222332455004452
t+1
t

4 6
2 10 .
5 1
Let now (3.4) be true for a fixed t > 1. Then T 5·11
= (T 5·11 )11 ≡ A11 (mod 11)
and it suffices to prove that A11 ≡ A (mod 11). Since A2 ≡ E (mod 11), we have
A2t ≡ (A2 )t ≡ E t ≡ E (mod 11) for any t ∈ N . Consequently, A11 ≡ A (mod 11),
which proves 3.5.
384
t
Lemma 3.6. For any t ∈ N we have det(T 5·11 − E) ≡ 0 (mod 11t+1 ).
P r o o f . If t = 1, then
det(T 55 − E) = 2 · 112 · 397 · 3742083511 ≡ 0 (mod 112 ).
t+1
Let the assertion be true for a fixed t > 1. First, it is evident that T 5·11
be written as
(3.5)
t+1
T 5·11
t
t
t
− E can
t
− E = (T 5·11 − E) · (E + T 5·11 + T 2·5·11 + . . . + T 10·5·11 ).
Now it follows from the induction hypothesis, from (3.5) and from Cauchy’s theorem
that it suffices to prove that
t
t
t
det(E + T 5·11 + T 2·5·11 + . . . + T 10·5·11 ) ≡ 0 (mod 11).
From (3.4) it follows that
t
t
t
E + T 5·11 + T 2·5·11 + . . . + T 10·5·11 ≡ E + A + A2 + . . . + A10 ≡ 6E + 5A (mod 11).
As congruent matrices have congruent determinants, we have
t
t
t
det(E + T 5·11 + T 2·5·11 + . . . + T 10·5·11 ) ≡ det(6E + 5A) = 132 ≡ 0 (mod 11).
This proves 3.6.
Theorem 3.7. For any t ∈ N , the system of congruences
(3.6)
t
(T 5·11 − E)x ≡ 0 (mod 11t+1 )
has exactly 11t+1 solutions and the number of solutions satisfying x 6≡ 0 (mod 11) is
equal to 10 · 11t . Moreover, if αt+1 is a solution of g(x) ≡ 0 (mod 11t+1 ), then each
solution of (3.6) can be expressed as [q, qαt+1 , qα2t+1 ], where q ∈ Z.
t
Put W = T 5·11 − E (mod 11t+1 ). From (3.4) it follows that all
6
entries of W , except for w33 , are units of the ring Z/11t+1 Z. Since 11 - det
6
there are coefficients r, s that are also units of the ring Z/11t+1 Z, for which
P r o o f.
the
4
,
1
r(w11 , w12 ) + s(w21 , w22 ) ≡ (w31 , w32 ) (mod 11t+1 ).
385
Thus there is a linear combination of the first and second rows of W transforming
W x ≡ 0 (mod 11t+1 ) to an equivalent form
(3.7)
w11
 w21
0

w12
w22
0
   
0
w13
a
w23   b  ≡  0  (mod 11t+1 ).
′
0
c
w33
′
Let us now prove that w33
≡ 0 (mod 11t+1 ). Multiplying the first row in (3.7) by a
suitable unit and, subsequently, adding it to the second row yields
(3.8)
w11
 0
0

w12
′
w22
0
   
a
w13
0
′  

b ≡ 0  (mod 11t+1 ).
w23
′
c
0
w33
′
′
The determinant of the matrix of (3.8) is w11 w22
w33
and, by Lemma 3.6, we have
′
′
t+1
′
w11 w22 w33 ≡ 0 (mod 11 ). Now it follows from (3.4) that w11 and w22
are units
t+1
′
t+1
of Z/p Z and thus w33 ≡ 0 (mod 11 ). This implies that the system W x ≡ 0
(mod 11t+1 ) is equivalent to the system
(3.9)
w11 a + w12 b + w13 c ≡ 0 (mod 11t+1 ),
w21 a + w22 b + w23 c ≡ 0 (mod 11t+1 ),
in which all the coefficients are units of Z/pt+1 Z. As no subdeterminant of the
system matrix of (3.9) is divisible by 11, any of the unknowns a, b, c can be chosen as
a parameter to express the other unknowns in a unique manner. Thus, each solution
of W x ≡ 0 (mod 11t+1 ) can be written as [qu1 , qu2 , qu3 ] for a fixed triple of units
u1 , u2 , u3 and a parameter q ∈ Z. Therefore the number of non-congruent solutions
to (3.6) is equal to the number of elements of the ring Z/11t+1 Z, which is 11t+1 , and
the number of solutions of the form x 6≡ 0 (mod 11) is equal to the number of units
of this ring, which is 10 · 11t .
Let us now prove that the solutions to (3.6) are exactly the triples [q, qαt+1 , qα2t+1 ]
where q ∈ Z. As the number of non-congruent triples [q, qαt+1 , qα2t+1 ] is equal
to 11t+1 , it suffices to show that h(11t+1 )[q, qαt+1 , qα2t+1 ] | 5 · 11t . As α = 9 is
a simple root of g(x) ≡ 0 (mod 11), we obtain by Hensel’s lemma, that for each
t ∈ N there is αt , which is uniquely determined modulo 11t , satisfying g(x) ≡
0 (mod 11t ) and such that α1 = α and αt ≡ αt−1 (mod 11t−1 ). Let ord11t (ε) for
ε 6≡ 0 (mod 11) denote the order of ε in the multiplicative group of Z/11t Z. Clearly,
2
h(11t+1 )[q, qαt+1 , qαt+1
] = ord11t+1 (αt+1 ) for any q ∈ Z where q 6≡ 0 (mod 11).
From ord11 (α1 ) = 5 and αt+1 ≡ α1 (mod 11) it now follows α5t+1 ≡ 1 (mod 11) for
5·11t
≡ 1 (mod 11t+1 ). Hence ord11t+1 (αt+1 ) | 5 · 11t .
any t ∈ N and thus αt+1
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According to Theorem 3.7, the set of all non-congruent solutions to (3.6) can be
written as E(αt+1 ) = {[q, qαt+1 , qα2t+1 ], q ∈ Z/pt+1 Z} and viewed as the eigenspace
associated with the eigenvalue αt+1 .
R e m a r k 3.8. The equality ord11t (αt ) = 5 · 11t−1 is a non-trivial consequence of
3.3 and 3.7 for each t ∈ N . See also Lemma 4.6 in [1].
Lemma 3.9. There exists no triple [a, b, c] for which h(11t )[a, b, c] = 5 · 11t .
t−1
P r o o f . It suffices to prove that the systems (T 5·11 − E)x ≡ 0 (mod 11t ) and
t
(T 5·11 − E)x ≡ 0 (mod 11t ) have identical solution sets for any t > 1. Denote by
t−1
X the set of all solutions of (T 5·11
− E)x ≡ 0 (mod 11t ) and by Y the set of all
t
solutions of (T 5·11 − E)x ≡ 0 (mod 11t ). The inclusion X ⊆ Y follows immediately
from the equality
t
t−1
T 5·11 − E = (E + T 5·11
t−1
+ T 2·5·11
t−1
+ . . . + T 10·5·11
t−1
) · (T 5·11
− E).
t
Modifying the proof of 3.7, we can determine that (T 5·11 − E)x ≡ 0 (mod 11t ) has
t−1
11t solutions, thus the same number as (T 5·11 − E)x ≡ 0 (mod 11t ). The equality
of the sets X and Y follows from their finiteness.
Now we can summarize our results in the main theorem:
Theorem 3.10. For any triple [a, b, c] 6≡ [0, 0, 0] (mod 11), we have:
If [a, b, c] 6∈ E(αt ) and c ≡ 3a + 5b (mod 11), then h(11t )[a, b, c] = 10 · 11t−1 .
If [a, b, c] 6∈ E(αt ) and c 6≡ 3a + 5b (mod 11), then h(11t )[a, b, c] = 10 · 11t .
If [a, b, c] ∈ E(αt ), then h(11t )[a, b, c] = ord11t (αt ) = 5 · 11t−1 .
References
t
[1] J. Klaška: Tribonacci modulo p . Math. Bohem. 133 (2008), 267–288.
[2] A. Vince: Period of a linear recurrence. Acta Arith. 39 (1981), 303–311.
zbl
[3] M. E. Waddill: Some properties of a generalized Fibonacci sequence modulo m. The
Fibonacci Quarterly 16 (Aug. 1978), 344–353.
zbl
Author’s address: Jiří Klaška, Department of Mathematics, Brno University of Technology, Technická 2, 616 69 Brno, Czech Republic, e-mail: klaska@fme.vutbr.cz.
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